How find this equation integer solution:$x^2y^2=4x^5+y^3$

find $x,y\in Z$,and such that $$x^2y^2=4x^5+y^3$$

and I use mathmatical give me:

my try:

Let $gcd(x,y)=d$ and put $x=da, y=db$, with $a$ and $b$ coprime:

$d^4a^2b^2 = 4d^5a^5 + d^3b^3 \Rightarrow da^2b^2=4d^2a^5+b^3$. Hence, we have that $a|b^3$ and as $gcd(a,b)=1$, $a=1$ or $a=-1$. Plugging it back, we have $4d^2-db^2+b^3=0$ or $4d^2+db^2-b^3=0$.

The discriminant, in the first case, is $b^4-16b^3$, which must be a perfect square. Therefore, $b^2-16b = k^2$, where $k$ is an integer. This gives us $(b-8)^2 - k^2 = 64$.

From this, we get the following pairs: $(125,3025), (27, 486), (54, 972), (32, 512)$.

In the second case, the discriminant is $b^4+16b^3$, which must be a perfect square again. Therefore, $b^2+16b=l^2$, where $l$ is an integer. This gives us $(b+8)^2-l^2 = 64$.

From this, we get the following pairs: $(0,0), (2,-4), (-1,2), (27, -243)$

my try is true? and have other methods?

Here's another approach:

$$x^2 y^2 = 4 x^5 + y^3$$

This is trivially satisfied by for $x=0, y=0$ and $x=0 \Leftrightarrow y = 0$

Now assume $x,y \ne 0$

For an odd prime $p$ suppose $p^n \parallel x$, and $p^m \parallel y$, with $x = p^n \tilde{x}$ and $y = p^m \tilde{y}$, so that $p \nmid \tilde{x}$ and $p \nmid \tilde{y}$

Then we have

$$p^{2(n+m)} \tilde{x}^2 \tilde{y}^2 = 4 p^{5n} \tilde{x}^5 + p^{3m} \tilde{y}^3$$

Now if $m > \frac{5}{3}n$

$$3m - 5n > 3 \frac{5}{3}n - 5n = 0$$ $$2(n+m) - 5n > 2(n+\frac{5}{3}n) - 5n = \frac{1}{3}n \ge 0$$

and

$$ p^{2(n+m)-5n} \tilde{x}^2 \tilde{y}^2 = 4 \tilde{x}^5 + p^{3m-5n} \tilde{y}^3 \Rightarrow p | \tilde{x}$$

And if $n > \frac{3}{5}m$

$$5n - 3m > 5 \frac{3}{5}m - 3m = 0$$

$$2(n+m) - 3m > 2(\frac{3}{5}m+m) -3m = \frac{1}{5}m \ge 0$$

and

$$ p^{2(n+m)-3m} \tilde{x}^2 \tilde{y}^2 = 4 p^{5n-3m} \tilde{x}^5 + \tilde{y}^3 \Rightarrow p | \tilde{y}$$

Hence we must have $3m = 5n$, and $n = 3k$, $m = 5k$ for some $k \in \mathbb{N}_0$.

So for each odd prime, p, $p^{3k} \parallel x$ and $p^{5k} \parallel y$ for some $k \in \mathbb{N}_0$. It follows that $x,y$ are of the form

$$x= 2^a z^3, y = \pm 2^b z^5$$ where $a,b \in \mathbb{N}_0, z\in \mathbb{Z}$ is an odd integer. (Note the sign of $x$ is determined by $z$)

Now we have

$$2^{2(a+b)} z^{16} = 2^{5a+2} z^{15}\pm 2^{3b} z^{15}$$

And with the trivial case $z=0$ excluded

$$2^{2(a+b)} z = 2^{5a+2} \pm 2^{3b} $$

Now if $b > \frac{5a + 3}{3}$ then

$$3b - (5a+2) > 3 \frac{5a + 3}{3} -(5a+2) = 1 > 0$$ $$2(a+b) -(5a+2) > 2(a+\frac{5a+3}{3}) -(5a+2) = \frac{1}{3}a \ge 0$$ and $$2^{2(a+b)-(5a+2)} z = 1 \pm 2^{3b-(5a+2)} \Rightarrow 2 | 1$$

Now if $a < \frac{3a}{5}b$ then

$$5a+2 - 3b > 5 \frac{3}{5}b - 3b = 1 > 0$$ $$2(a+b) - 3b > 2(\frac{3b}{5}+b) -3b = \frac{1}{5}b \ge 0$$ and

$$2^{2(a+b)-3b} z = 2^{5a+2-3b} \pm 1 \Rightarrow 2 | 1$$

Hence we must have $5a \le 3b \le 5a+3$

Let $c = 3b - 5a, c=0,1,2,3$ then

$$2(a+b) = 2(a+ \frac{5a + c}{3}) = \frac{16a + 2c}{3}$$

$$2^{(16a + 2c)/3} z = 2^{5a+2} \pm 2^{5a+c} $$

$$2^{(a + 2c)/3} z = 4 \pm 2^{c} $$

For $c=0$,

$2^{a/3} z = 4 \pm 1 \Rightarrow a=0, b=0, z=3,5$

$(x,y)=(2^0 3^3,- 2^0 3^5), (2^0 5^3, 2^0 5^5) = (27,-243), (125,3125)$

For $c=1$,

$2^{(a+2)/3} z = 4 \pm 2 \Rightarrow a=1, b=2, z=1,3$

$(x,y)=(2^1 1^3,-2^2 1^5), (2^1 3^3,2^2 3^5)=(2,-4), (54,972)$

For $c=2$,

$2^{(a+4)/3} z = 4 \pm 4 \Rightarrow a=5, b=9, z=0,1$

$(x,y)=(2^5 0^3,-2^9 0^5), (2^5 1^3,2^9 1^5)=(0,0), (32,512)$

For $c=3$,

$2^{(a+6)/3} z = 4 \pm 8 \Rightarrow a=0, b=1, z=-1,3$

$(x,y)=(2^0(-1)^3,-2^1(-1)^5), (2^0 3^3,2^1 3^5)=(-1,2), (27,486)$

This is a more generalizable method than the one above which requires that the combined powers of $x$ and $y$ in each term cover no larger a range than $2$ in order to obtain a quadratic in $d$. For instance, a near identical treatment solves $x^2 y^6 = 2^5 x^{13} + y^7$.

The basic equation is defined here for real $(x,y)$ as well, by: $$ x^2 y^2 = 4 x^5 + y^3 $$ See picture ; window $-200 < x < +200$ , $-4000 < y < +4000$ .
It is immediately clear that $(x,y) = (0,0)$ is an (integer) solution of the equation and that $x = 0 \leftrightarrow y = 0$ . $\color{red}{Tangents}\,$ at the curve are calculated by implicit differentation: $$ 2 x y^2 + x^2 2 y y' = 20 x^4 + 3 y^2 y' \quad \Longrightarrow \quad y' = \frac{20 x^4 - 2 x y^2}{2 y x^2 - 3 y^2} $$ Horizontal tangents at: $$ 20 x^4 - 2 x y^2 = 0 \quad \Longrightarrow \quad y^2 = 10 x^3 $$ Substitute into the basic equation to obtain: $$ x^2 y^2 = 4 x^5 + y^3 \quad \Longrightarrow \quad x^2 \cdot 10 x^3 = 4 x^5 + 10 x^3 y $$ We have covered the trivial solution $(0,0)$ so forget about it in the sequel. $$ 6 x^2 = 10 y \quad \Longrightarrow \quad 36 x^4 = 100 y^2 = 1000 x^3 \quad \Longrightarrow \quad (x_S,y_S) = \left(\frac{10^3}{6^2},\sqrt{\frac{10^{10}}{6^6}}\right) $$ The Special point  S  is important for the numerical work that follows. Vertical tangents at: $$ 2 y x^2 - 3 y^2 = 0 \quad \Longrightarrow \quad y = \frac{2}{3} x^2 $$ Substitute into the basic equation to obtain: $$ x^2 \cdot \frac{4}{9} x^4 = 4 x^5 + \frac{8}{27} x^6 \quad \Longrightarrow \quad \frac{4}{27} x = 4 \quad \Longrightarrow \quad (x,y) = (27,486) $$ The latter happens to be one of the required integer solutions !
The rest of the method is brute force.
It is clear from the picture that there are four branches:

• one extending from $(0,0)$ towards minus infinity in the quadrant $x > 0$ , $y < 0$
  indicated as A
• one extending from $(0,0)$ towards plus infinity in the quadrant $x < 0$ , $y > 0$
  indicated as B
• one extending from the special point  S  towards plus infinity on the right side
  in the quadrant $x > 0$ , $y > 0$ , indicated as C
• one extending from the special point  S  towards plus infinity on the left side
  in the quadrant $x > 0$ , $y > 0$ , indicated as D

program krabbel;

function pow(x : double; n : integer) : double;
var
  q : double;
  k : integer;
begin
  q := 1;
  for k := 1 to n do
    q := q * x;
  pow := q;
end;

function f(x,y : double) : double;
begin
  f := pow(x,2)*pow(y,2) - (4*pow(x,5) + pow(y,3));
end;

procedure crawl_A;
var
  x,y : integer;
  g : double;
begin
  x := 0; y := 0;
  while true do
  begin
    y := y - 1;
    g := f(x,y);
    if g = 0 then Writeln(x,' ',y);
    while g > 0 do
    begin
      x := x + 1;
      g := f(x,y);
      if g = 0 then Writeln(x,' ',y);
    end;
  end;
end;

procedure crawl_B;
var
  x,y : integer;
  g : double;
begin
  x := 0; y := 0;
  while true do
  begin
    y := y + 1;
    g := f(x,y);
    if g = 0 then Writeln(x,' ',y);
    while g < 0 do
    begin
      x := x - 1;
      g := f(x,y);
      if g = 0 then Writeln(x,' ',y);
    end;
  end;
end;

procedure crawl_C;
var
  x,y : integer;
  g : double;
begin
  x := Round(pow(10,3)/pow(6,2));
  y := Round(sqrt(pow(10,10)/pow(6,6)));
  while true do
  begin
    y := y + 1;
    g := f(x,y);
    if g = 0 then Writeln(x,' ',y);
    while g > 0 do
    begin
      x := x + 1;
      g := f(x,y);
      if g = 0 then Writeln(x,' ',y);
    end;
  end;
end;

procedure crawl_D;
var
  x,y : integer;
  g : double;
begin
  x := 27; y := 486;
  while true do
  begin
    y := y + 1;
    g := f(x,y);
    if g = 0 then Writeln(x,' ',y);
    while g < 0 do
    begin
      x := x + 1;
      g := f(x,y);
      if g = 0 then Writeln(x,' ',y);
    end;
  end;
end;

begin
{ crawl_A; crawl_B; } crawl_C; { crawl_D; }
end.

Results $\;\rightarrow picture: \color{red}{dots}$

by hand:
0 0
27 486

crawl_A:
2 -4
27 -243

crawl_B:
-1 2

crawl_C:
32 512
54 972
125 3125

crawl_D:
nothing